A common analysis scheme for determind the calcium content of a compound involves the precipitation of Ca2+ ions as CaC2O4 (calcium oxalate). This precipitate is appreciable soluble in water. CaC2O4(s)<------->Ca2+(aq) + C2O42-(aq) Yet it can be safely washed with an aqueous solution of Na2C2O4 (sodium oxalate) without risk of loss. Why is this?
2007-02-24 10:02:17 · 3 answers · asked by calculusgirl1979 2 in Science & Mathematics ➔ Chemistry
CaC2O4(s) --------> Ca2+(aq) + oxalate 2- (aq)
You are adding oxalate forcing the above reaction to the left. As long as the molarity of oxalate in the wash is greater than solubility of the calcium oxalate you should be OK.
In fact, the way you precipitate out the calcium oxalate is to mix the calcium solution with a concentrated solution of oxalate to begin with.
2007-02-24 10:15:08 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Loss of calcium would only occur through the production of calcium hydroxide. However Calcium hydroxide is very insoluble in water so in the presence of the sodium ions produced from the sodium oxalate solution, sodium hydroxide is strongly favoured over calcium hydroxide. The cacium ions will stay in the equilibrium you've drawn in your question because Calcium hydroxide is so relatively unfavoured in comparison to calcium oxalate in the prescence of sodium ions.
2007-02-24 18:27:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Look up the "Common Ion Effect". That will explain this.
2007-02-24 18:06:54 · answer #3 · answered by Gervald F 7 · 0⤊ 0⤋